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Determinants — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDeterminants1 Mark
Question
Fill in the blanks:
$\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}=$ ________.
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Answer

$\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}$ $=(\text{z}-\text{x})(\text{y}-\text{z})(\text{y}-\text{x}+\text{xyz})$
Solution:
We have, $\begin{vmatrix}0&\text{xyz}&\text{x}-\text{z}\\\text{y}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}=\begin{vmatrix}\text{z}-\text{x}&\text{xyz}&\text{x}-\text{z}\\\text{z}-\text{x}&0&\text{y}-\text{z}\\\text{z}-\text{x}&\text{z}-\text{y}&0\end{vmatrix}$ $\big[\because\ \text{C}_1\rightarrow\text{C}_1-\text{C}_3\big]$
$=(\text{z}-\text{x})\begin{vmatrix}1&\text{xyz}&\text{x}-\text{z}\\1&0&\text{y}-\text{z}\\1&\text{z}-\text{y}&0\end{vmatrix}$
[Taking (z - x) common from column 1]
Expanding along $R_1,$
$= (\text{z} - \text{x})\Big[1.\big\{-1({\text{y} - \text{z})(\text{z} - \text{y})}\big\} -\text{xyz}(\text{z} - \text{y}) + (\text{x} - \text{z})(\text{z} - \text{y})\Big]$
$=(\text{z}-\text{x})(\text{z}-\text{y})(-\text{y}+\text{z}-\text{xyz}+\text{x}-\text{z})$
$=(\text{z}-\text{x})(\text{z}-\text{y})(\text{x}-\text{y}-\text{xyz})$
$=(\text{z}-\text{x})(\text{y}-\text{z})(\text{y}-\text{x}+\text{xyz})$

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